Does this integral converge $\int_{-\infty}^{\infty} \frac{e^{-x}}{1+x^2}\,dx$

Question

I need to check whether this integral converges or not $\int^{\infty}_{-\infty} \frac{e^{-x}}{1+x^2}\,dx$

I substituted $y=-x$ then this integral transformed to $\int^{\infty}_{-\infty} \frac{e^{y}}{1+y^2}\,dy$ , then I thought of dividing it into two parts from $-\infty$ to $0$ and then from $0$ to $\infty$, in the first case I think area will be finite but in the second case it's not since $e^x$ grows rapidly, so it diverges!

I could not follow your statement "I substituted $y=−x$ then this integral transformed to $\int \frac{e^y} {1+y^2}dx$ ," — , Nov 21 '13 at 14:02
It doesn't converge. For $x\to -\infty$, the factor $e^{-x}$ grows exponentially (well, how else ;). — Daniel Fischer, Nov 21 '13 at 14:05
Have you a course of complex analysis Neeraj? You may try it using calculus of residue. — Supriyo, Nov 21 '13 at 14:39
TIFR GS2012 Prob. 24. I shall try to write an easier solution. — Supriyo, Nov 23 '13 at 04:29

score 1 · Answer 1 · answered Nov 21 '13 at 14:15

1

Hint: Check $\lim_{x\to\pm\infty}\frac{e^{-x}}{1+x^2}$. Are both zero? If not, then this implies that the integral does not converge (since the integrand is strictly positive, we can't have cancellation as in the $\sin(x^2)$ case).

answered Nov 21 '13 at 14:15

Clayton

24,751

1

To be completely clear, the integrand doesn't necessarily have to converge to 0, but if it converges to something, it must be zero for the integral to converge. – Najib Idrissi Nov 21 '13 at 14:17
@egreg: I may be mistaken, but no it's not. Consider a function that makes increasingly narrow triangles but with constant height. – Najib Idrissi Nov 21 '13 at 14:21
@nik: You are correct. Thank you for making it more explicit. Fortunately, here both do converge (in the extended real numbers sense). – Clayton Nov 21 '13 at 14:23
@nik I think you're right; the correct argument should be that this function is monotone in a neighborhood of $\infty$ and of $-\infty$. – egreg Nov 21 '13 at 14:24
@Clayton: Yes, your answer is fine; I'm just making clear all the details for the OP. – Najib Idrissi Nov 21 '13 at 14:24
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@egreg Note that there are positive unbounded functions $f$ such that $$\int f(x),dx<\infty $$ Consider a "sum of spikes" function such that $f(n)=n$ on while small otherwise. – AD - Stop Putin - Nov 21 '13 at 14:26
@AD. Yes, I realized that and removed my comment. – egreg Nov 21 '13 at 14:28

score 0 · Answer 2 · answered Nov 21 '13 at 14:06

0

Hint : You first calculate $$\int_a^b \frac{e^{-x}}{1+x^2}dx$$

and then send $b\rightarrow \infty$ and $a\rightarrow -\infty$.

If this limit exists then, your integral converges.

answered Nov 21 '13 at 14:06

why negative vote :O – Nov 21 '13 at 14:23
I am not the downvoter, but calculating that integral is hard. Deciding convergence is easy. In other words this answer is no real help. – AD - Stop Putin - Nov 21 '13 at 14:32
@AD. Oh.. yes... I could not realize that before.. Thank you for mentioning that.. :) – Nov 21 '13 at 14:34

Does this integral converge $\int_{-\infty}^{\infty} \frac{e^{-x}}{1+x^2}\,dx$

2 Answers2